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Chapter 8: Problem 9

A random sample of 25 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 10 and the sample standard deviation is \(2 .\) Use a level of significance of \(0.05\) to conduct a two-tailed test of the claim that the population mean is \(9.5\). (a) Cbeck Requirements Is it appropriate to use a Student's \(t\) distribution? Explain. How many degrees of freedom do we use? (b) What are the hypotheses? (c) Compute the sample test statistic \(t .\) (d) Estimate the \(P\) -value for the test. (e) Do we reject or fail to reject \(H_{0} ?\) (f) Interpret the results.

Short Answer

Expert verified
We fail to reject the null hypothesis; there isn't enough evidence to claim the population mean differs from 9.5.

Step by step solution

01

Checking Requirements

To determine if it is appropriate to use a Student's \( t \) distribution, recall that it can be used if the population is normally distributed, which the problem states as mound-shaped and symmetric, and/or the sample size is less than 30. Given our sample size of 25, it is reasonable to use the \( t \) distribution. With a sample size of 25, the degrees of freedom (df) are calculated as \( n - 1 = 25 - 1 = 24 \).
02

Defining Hypotheses

The null hypothesis \( H_0 \) states that the population mean \( \mu \) equals 9.5, and the alternative hypothesis \( H_a \) states that the population mean \( \mu \) is not equal to 9.5. Thus, we have:- Null Hypothesis \( H_0: \mu = 9.5 \)- Alternative Hypothesis \( H_a: \mu eq 9.5 \)
03

Computing the Test Statistic

Calculate the test statistic \( t \) using the formula \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} = 10 \), \( \mu = 9.5 \), \( s = 2 \), and \( n = 25 \). Substituting the values gives:\[t = \frac{10 - 9.5}{2/\sqrt{25}} = \frac{0.5}{0.4} = 1.25.\]
04

Estimating the P-value

To estimate the \( P \)-value, use the \( t \)-distribution table or software for \( t = 1.25 \) with 24 degrees of freedom. For a two-tailed test, find the probabilities of \( t \) being greater than 1.25 or less than -1.25. This cumulative probability corresponds to a \( P \)-value between 0.1 and 0.2.
05

Making a Decision

Given the \( P \)-value (between 0.1 and 0.2) and the significance level of \( \alpha = 0.05 \), we see that the \( P \)-value is greater than \( \alpha \). Therefore, we fail to reject the null hypothesis \( H_0 \).
06

Interpreting the Results

Since we fail to reject the null hypothesis, there is not enough statistical evidence to conclude that the population mean differs from 9.5 at the 0.05 significance level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Student's t-distribution
The Student's t-distribution is essential for conducting hypothesis tests, especially when the sample size is small, typically less than 30. It is similar to the normal distribution but has thicker tails, which means it is more prone to capturing the spread of data more accurately under conditions of smaller sample sizes. This characteristic allows for better estimation when uncertainty is high due to a smaller dataset.
In the case of the exercise, we use the t-distribution because:
  • The sample size is 25, which is indeed less than 30.
  • The data is drawn from a symmetric and mound-shaped distribution.
These two considerations justify employing the t-distribution over the standard normal distribution. It helps account for the extra variability expected when dealing with small sample sizes, making it a pivotal tool in inferential statistics.
Sample Standard Deviation
The sample standard deviation is a measure of how spread out the values in a sample are around the mean.It is an important component in calculating the test statistic in a t-test, as it reflects the variability in the sample data.
The formula for the sample standard deviation is: \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]where:
  • \( s \) is the sample standard deviation.
  • \( n \) is the sample size.
  • \( x_i \) represents each value in the sample.
  • \( \bar{x} \) is the sample mean.

In our specific exercise, the calculated sample standard deviation is 2.This value is used to compute the test statistic, indicating how the means of repeated samples might differ from the true population mean.
Degrees of Freedom
Degrees of freedom (df) refer to the number of values in a calculation that are free to vary.In the context of the t-distribution, it influences the shape of the distribution, adjusting how extreme or conservative the t-distribution will be.
The formula to find the degrees of freedom in a t-test is: \[ df = n - 1 \]where \( n \) is the sample size.For our exercise, this calculation yielded:\[ df = 25 - 1 = 24 \]
Understanding the degrees of freedom helps in selecting the correct t-distribution during table look-up or when estimating critical values in hypothesis testing.More degrees of freedom result in a distribution that more closely resembles a normal distribution, which is why the condition \( n-1 = 24 \) confirms the appropriateness of using the t-distribution in our hypothesis testing.

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Most popular questions from this chapter

Let \(x\) be a random variable that represents the \(\mathrm{pH}\) of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the \(x\) distribution is \(\mu=7.4\) (Reference: Merck Manual, a commonly used reference in medical schools and nursing programs). A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that \(\bar{x}=8.1\) with sample standard deviation \(s=1.9 .\) Use a \(5 \%\) level of significance to test the claim that the drug has changed (either way) the mean \(\mathrm{pH}\) level of the blood.

For a random sample of 20 data pairs, the sample mean of the differences was \(2 .\) The sample standard deviation of the differences was \(5 .\) Assume that the distribution of the differences is mound-shaped and symmetric. At the \(1 \%\) level of significance, test the claim that the population mean of the differences is positive. (a) Is it appropriate to use a Student's \(t\) distribution for the sample test statistic? Explain. What degrees of freedom are used? (b) State the hypotheses. (c) Compute the sample test statistic. (d) Estimate the \(P\) -value of the sample test statistic. (e) Do we reject or fail to reject the null hypothesis? Explain. (f) What do your results tell you?

Prose rhythm is characterized by the occurrence of five-syllable sequences in long passages of text. This characterization may be used to assess the similarity among passages of text and sometimes the identity of authors. The following information is based on an article by D. Wishart and S. V. Leach appearing in Computer Studies of the Humamities and Verbal Behavior (Vol. 3, pp. 90-99). Syllables were categorized as long or short. On analyzing Plato's Republic, Wishart and Leach found that about \(26.1 \%\) of the five-syllable sequences are of the type in which two are short and three are long. Suppose that Greek archaeologists have found an ancient manuscript dating back to Plato's time (about \(427-347\) B.C. \() .\) A random sample of 317 five-syllable sequences from the newly discovered manuscript showed that 61 are of the type two short and three long. Do the data indicate that the population proportion of this type of five-syllable sequence is different (either way) from the text of Plato's Republic? Use \(\alpha=0.01\).

Education Education influences attitude and lifestyle. Differences in education are a big factor in the "generation gap." Is the younger generation really better educated? Large surveys of people age 65 and older were taken in \(n_{1}=32\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{1}=15.2 \%\) of the older adults had attended college. Large surveys of young adults (ages \(25-34)\) were taken in \(n_{2}=35\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{2}=19.7 \%\) of the young adults had attended college. From previous studies, it is known that \(\sigma_{1}=7.2 \%\) and \(\sigma_{2}=5.2 \%\) (Reference: American Generations by S. Mitchell). Does this information indicate that the population mean percentage of young adults who attended college is higher? Use \(\alpha=0.05\).

A random sample of 49 measurements from a population with population standard deviation 3 had a sample mean of 10\. An independent random sample of 64 measurements from a second population with population standard deviation 4 had a sample mean of \(12 .\) Test the claim that the population means are different. Use level of significance \(0.01\). (a) What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute \(\bar{x}_{1}-\bar{x}_{2}\) and the corresponding sample test statistic. (d) Find the \(P\) -value of the sample test statistic. (e) Conclude the test. (f) The results.
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